The time the truck must apply the given force to increase its speed to given value is 5 s.
The given parameters;
- <em>applied force, F = 600 N</em>
- <em>mass of the truck, m = 1,500 kg</em>
- <em>speed of the truck, v = 2 m/s</em>
The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>
F = ma

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.
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Respuesta:28,64 m/s.
Explicación:Datos:
Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²
El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.
Entonces tenemos que:

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.
Sustituyendo tenemos que:

Que tengas un buen día!
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
Answer:
- Particles smaller than atoms are called subatomic particles .
- There are three famous subatomic particles, proton, neutron and electron .
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A. 10 rations = 1 deca-ration.
b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.
c. 10⁻⁵ phones = 1 micro-phones.
d. 10⁻⁹ goats = 1 nano-goats.
e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.